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This previous post is the closest thing to an answer I could find, but I don't quite understand why the load will still see the full voltage of the source and not the zener voltage. The contributor who answered stated that since the combination of zener and series resistor is in parallel with the load, the load will still see the full voltage of the source. What I'm having a hard time understanding is how the resistor placement affects the voltage seen by the load. Thank you in advance

EDIT: Here is the circuit in question (from the post I linked to) enter image description here

EDIT 2: Okay, I'm slowly realizing my faults in asking a question without providing as much specific detail as possible. Here goes:

What I'm trying to do: I have a bicycle hub dynamo that produces single phase AC voltage that can reach up to 60V peak (120V pk-pk) going about 55mph.

I'm trying to rectify and regulate the voltage down to 5V so I can charge a cellphone. For the most part I have the 5V regulation down (I'm using a DC-DC buck converter (TI's LM2596).

The issue I'm trying to resolve is during the rectification stage. I'm using a bridge rectifier and smoothing capacitor to rectify and smooth the voltage coming out of the dynamo. However, the smoothing capacitor I have is only rated at 35V so at those higher outputs of the dynamo (ie. above 35V peak), I'm assuming I would blow the smoothing capacitor up without a zener voltage regulator.

Here is a more accurate picture of my setup (I simulated the dynamo using a transformer at the input):

enter image description here

So in the picture, I have placed the series resistor correctly. My original question is: why can't I put the series resistor after the zener diode?

EDIT 3: For completeness here is the simulation for the incorrect schematic. The zener has 15V breakdown, but the output still sees the rectified ~30V of the source: enter image description here

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  • \$\begingroup\$ You should use the other schematic in the answer. \$\endgroup\$
    – crossroad
    Apr 28 at 19:26
  • \$\begingroup\$ The problem with the linked question is that the accepted answer did not include a full, correct schematic, so you only have an incomplete correct schematic or a full incorrect one. I think I will go edit that answer... \$\endgroup\$
    – Theodore
    Apr 28 at 19:26
  • \$\begingroup\$ @crossroad the other schematic doesn't show the LED or its dropping resistor. \$\endgroup\$
    – Theodore
    Apr 28 at 19:27
  • \$\begingroup\$ Apologies: I edited the post to include the picture I was referring to (as in why does this circuit not regulate voltage down to the zener voltage). I didn't review the picture to ensure that the source voltage is higher than the zener voltage, I'm mainly interested in the setup and why in any case it will not work. \$\endgroup\$ Apr 28 at 19:28
  • \$\begingroup\$ @Theodore From my understanding, the poster here is not interested specifically in LEDs. I think a generic load would suffice. \$\endgroup\$
    – crossroad
    Apr 28 at 19:28
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) or (b) will result in Zener regulation. (c) doesn't because the load is connected directly to the bridge rectifier.

In the arrangement of 1c R3 and D3 just apply a load across the rectified supply. You will have VZD across the Zener and the remainder of the supply voltage dropped across R3.

Bicycle alternators (technically they're not dynamos which are DC machines) have a high inductance. The impedance of an inductor is given by \$ Z = 2\pi fL \$ where \$f\$ is the frequency and \$L\$ the inductance. That means that the impedance is in series with the load and is low at low speed and increases with speed. Most manufacturers design the device so that the impedance coupled with the intended load tends to result in a more constant output voltage over a wide range of speeds. The old bulb systems would flicker at very low speed but at a moderate speed would illuminate well yet not burn out on the downhill run.

enter image description here

Figure 1. There is some very good information on Bicycle Electronics with many circuits and performance curves.

I recommend that you figure out how to measure your alternator's inductance and add that into your simulator. Just put it in series with a sine signal source.

Links:

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  • \$\begingroup\$ Thank you for the thorough response. This helped a bunch! \$\endgroup\$ Apr 29 at 1:34
  • \$\begingroup\$ @blackmcgraw, I've added four more links. Don't forget to upvote useful answers and accept one that answers your question. \$\endgroup\$
    – Transistor
    Apr 29 at 18:06
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The load is placed in parallel with the source, so there will be exactly the source voltage at the load. The zener voltage, whatever it may be, is over the zener only. The load would see the zener voltage only if the load was over the zener.

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  • \$\begingroup\$ Thank you very much for your input. This is a very common sense way that you explained it and I don't know why I struggled with the concept so hard! \$\endgroup\$ Apr 29 at 1:37
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If you connect the zener directly across the bridge rectifier then it will have to dissipate all excess energy to load the dynamo down to the zener voltage. With a series resistance, the resistor will dissipate some of the energy. Relatively high power resistors are cheap, readily available and reliable. The resistor also softens the switch-on of the zener, which may have EMC benefits. If a resistor is placed after the zener it serves no purpose.

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  • \$\begingroup\$ Thank you for your answer. I guess I'm having a hard time understanding why placing the resistor after the zener doesn't also dissipate some of the excess energy? Once the zener "turns on" then current flows through the resistor right? \$\endgroup\$ Apr 28 at 20:05
  • \$\begingroup\$ @blackmcgraw When you say 'after' do you mean between the zener and the negative rail or between the zener and the capacitor (and therefore the load). In the former case the load will see the zener voltage plus the voltage across the resistor, which is probably not what you want. In the latter case, the load will see the zener voltage minus the voltage drop across the resistor, which will depend on the resistance and the load current. \$\endgroup\$
    – Frog
    Apr 29 at 1:29
  • \$\begingroup\$ I meant the former case. This explanation helped clarify things for me thank you very much! \$\endgroup\$ Apr 29 at 1:34
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Looking at the first schematic in the linked post, the reason is that the resistor is what actually reduces the voltage. The zener diode conducts current when the voltage is high enough, it's this current that cause a voltage drop across the resistor, effectively limiting the voltage seen by the load.

The schematic you currently include in your question is basically incorrect.

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  • \$\begingroup\$ Thank you for the input! Much appreciated \$\endgroup\$ Apr 29 at 1:35
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In your simulation replace V2 with an AC current source (to represent the rotating permanent magnet).

Reduce the K of your transformer to 0.95 or lower to represent the gap in the generator.

After that your simulations should give more realistic results.

also measure the short-circuit current of your generator, that should be fairly constant.

then just buy enough LEDs (or resistors) to use up the available current and call it done.

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